How can I count the occurrences of each element in a vector in MATLAB?

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I would like to be able to return the count of occurences of each element in a vector.
For example if I have a vector:
x=[10 25 4 10 9 4 4]
I expect the result to be
y=[2 1 3 2 1 3 3].

採用された回答

MathWorks Support Team
MathWorks Support Team 2020 年 2 月 27 日
編集済み: MathWorks Support Team 2020 年 2 月 27 日
As of MATLAB R2019a, you can use the “groupcounts” function to compute the number of times an element appears in a vector as a summary. In other words, the elements of the below output “GC” are the counts of the corresponding element values in “GR” (from the original input vector “x”):
x = [10 25 4 10 9 4 4]';
[GC,GR] = groupcounts(x)
GC =
3
1
2
1
GR =
4
9
10
25
For more information on "groupcounts", please see the documentation link below:
---
As of MATLAB R2018b, you can use the “grouptransform” function if you want to compute the number of times an element appears in a vector and output that count for each corresponding element of the input vector. For example:
x = [10 25 4 10 9 4 4]';
y = grouptransform(x,x,@numel)
y =
2
1
3
2
1
3
3
For more information on "grouptransform", please see the documentation link below:
---
Prior to MATLAB R2018b, while there is no single function to count occurrences of each element, there are a few ways to count elements in a vector:
1. Logical Indexing:
The following code snippet will give the desired output:
y = zeros(size(x));
for i = 1:length(x)
y(i) = sum(x==x(i));
end
For MATLAB R2016b and later, you can use implicit expansion to further simplify the code:
y = sum(x==x')
2. Binning:
You can use the "hist" and "unique" functions as shown here to do the same:
x = [10 25 4 10 9 4 4]
[a,b]=hist(x,unique(x))
3. Third-Party Tools:
For another workaround, see the following file, 'CountMember.m', that was contributed by a MATLAB user to do the same from a single function:
Note that MathWorks does not guarantee or warrant the use or content of submissions to the MATLAB Central File Exchange. Any questions, issues, or complaints should be directed to the contributing author.
  3 件のコメント
Walter Roberson
Walter Roberson 2020 年 5 月 19 日
S = 'aaabb'
nnz(triu((S'=='a') & S=='b'))
Walter Roberson
Walter Roberson 2022 年 4 月 26 日
https://www.mathworks.com/help/matlab/ref/double.groupcounts.html#mw_92fbcf5a-2ab5-45d0-ac09-68c1986c269f
when you use groupcounts then the groups are in the order returned by unique('sorted')

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その他の回答 (3 件)

Andrei Bobrov
Andrei Bobrov 2014 年 8 月 14 日
[a,b] = histc(x,unique(x));
y = a(b);

Razvan Carbunescu
Razvan Carbunescu 2019 年 5 月 9 日
There is a simpler way of answering this now using groupcounts(R2019a) or grouptransform(R2018b):
>> x=[10 25 4 10 9 4 4]';
>> grouptransform(x,x,@numel)
ans =
2
1
3
2
1
3
3
>>[GC,GR]=groupcounts(x)
GC =
3
1
2
1
GR =
4
9
10
25
  4 件のコメント
Razvan Carbunescu
Razvan Carbunescu 2019 年 6 月 6 日
For the example you gave above how does the solution look and what does 'similar number' for the first column mean?
Razvan Carbunescu
Razvan Carbunescu 2019 年 6 月 7 日
This seems like a very different type of problem so unlikely the functions in this topic will help you directly. I'd post this question as a separate thread with the example input/output

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Julian Hapke
Julian Hapke 2017 年 6 月 1 日
編集済み: Julian Hapke 2017 年 6 月 1 日
here is another one:
sum(bsxfun(@eq,x,x'),1)
or if you want the output to be the same orientation as input
sum(bsxfun(@eq,x,x'),(size(x,2)==1)+1)
  1 件のコメント
Johannes Korsawe
Johannes Korsawe 2017 年 6 月 1 日
the second solution exhibits pathological tendencies...

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